Question

A cube of wood supporting $200gm$ mass just floats in water $(\rho =1g/cc)$. When the mass is removed, the cube is raised by $2cm$. The size of the cube is _____.

Answer 1

Let l be the side and ρ be the density of the cube.

Now, in initial condition

Weight of the cube + weight of mass placed above (200 gm) = weight of the water displaced

$l^3{\rho }_{wood}g+200g=l^3{\rho }_{w}g$

$l^3{\rho }_{wood}=l^3{\rho }_w-\mathrm{200....}\left(I\right)$

When the mass is removed

Then,

$l^3{\rho }_{wood}g=l^2(l-2)g....\left(II\right)$

Weight of the block = up thrust

Now, from equation (I) and (II)

$l^3{\rho }_w-200=l^2(l-2)$

We know that,

${\rho }_w=1$

So,

$l^3-200=l^3-2l^2$

$l^2=100$

$l=10cm$

Hence the sides of the cube is of 10 cm.